steph/papers/advanceshdl.pdf · advances in mathematics 204 (2006) 204–240 decomposable...

Advances in Mathematics 204 (2006) 204–240www.elsevier.com/locate/aim

Decomposable compositions, symmetricquasisymmetric functions and equality of ribbon

Schur functionsLouis J. Billeraa,1, Hugh Thomasb,∗,2,3, Stephanie van Willigenburgc,4

aDepartment of Mathematics, Cornell University, Ithaca, NY 14853-4201, USAbFields Institute, 222 College St., Toronto, Ont. Canada M5T 3J1

cDepartment of Mathematics, University of British Columbia, Vancouver, BC Canada V6T 1Z2Received 5 January 2005; accepted 18 May 2005

Communicated by Sara BilleyAvailable online 26 July 2005

Abstract

We define an equivalence relation on integer compositions and show that two ribbon Schurfunctions are identical if and only if their defining compositions are equivalent in this sense.This equivalence is completely determined by means of a factorization for compositions: equiv-alent compositions have factorizations that differ only by reversing some of the terms. As anapplication, we can derive identities on certain Littlewood–Richardson coefficients.

Finally, we consider the cone of symmetric functions having a nonnnegative representation in termsof the fundamental quasisymmetric basis. We show the Schur functions are among the extremes ofthis cone and conjecture its facets are in bijection with the equivalence classes of compositions.© 2005 Elsevier Inc. All rights reserved.

MSC: primary 05E05, 05A17; secondary 05A19, 05E10

Keywords: Ribbon Schur function; Littlewood–Richardson coefficients; Compositions; Partitions

∗ Corresponding author.E-mail addresses: [email protected] (L.J. Billera), [email protected] (H. Thomas),

[email protected] (S. van Willigenburg).1 Supported in part by NSF Grant DMS-0100323.2 Supported in part by the National Sciences and Engineering Research Council of Canada.3 Current address: Department of Mathematics and Statistics, University of New Brunswick, Fredericton,

NB Canada E3B 5A3.4 Supported in part by the National Sciences and Engineering Research Council of Canada.

0001-8708/$ - see front matter © 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.aim.2005.05.014

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L.J. Billera et al. / Advances in Mathematics 204 (2006) 204–240 205

1. Introduction

An important basis for the space of symmetric functions of degree n is the set ofclassical Schur functions s�, where � runs over all partitions of n. The skew Schur func-tions s�/� can be expressed in terms of these by means of the Littlewood–Richardson

coefficients c�� by

s�/� =∑�

c��s�. (1.1)

These coefficients also describe the structure constants in the algebra of symmetricfunctions. In particular they describe the multiplication rule for Schur functions,

s� s� =∑�

c��s�. (1.2)

From the perspective of the representation theory of the symmetric group, the coefficientc�� gives the multiplicity of the irreducible representation corresponding to the partition

� in the induced tensor product of those corresponding to � and �. In algebraic geometrythe c�

�� arise as intersection numbers in the Schubert calculus on a Grassmanian. Asa result of these and other instances in which they arise, the determination of thesecoefficients is a central problem.

We consider here the question of when two ribbon Schur functions might be equal.The more general question of equalities among all skew Schur functions has sincebeen broached in [14], where the question of when a skew Schur function can equal aSchur function is answered in the case of power series and their associated polynomials.The general question of equalities among skew Schur functions remains open. Any suchequalities imply equalities between certain pairs of Littlewood–Richardson coefficients.

In the case of ribbon Schur functions, that is skew Schur functions indexed by ashape known as a ribbon (or rim hook, or border strip; these are diagrams correspondingto connected skew shapes containing no 2 × 2 rectangle), we give necessary and suffi-cient conditions for equality. Ribbons are in natural correspondence with compositions,and equality arises from an equivalence relation on compositions, whose equivalenceclasses all have size equal to a power of two. This power corresponds to the num-ber of nonsymmetric compositions in a certain factorization of any of the underlyingcompositions in a class, and equivalence comes by means of reversal of terms.

A motivation for studying ribbon Schur functions is that they arise in various contexts.They were studied already by MacMahon [11, §168–169], who showed their coefficientsin terms of the monomial symmetric functions to count descents in permutations withrepeated elements. The scalar product of any two gives the number of permutationssuch that it and its inverse have the associated pair of descent sets [12, Corollary7.23.8]. They are also useful in computing the number of permutations with a givencycle structure and descent set [5]. Lascoux and Pragacz [9] give a determinant formulafor computing Schur functions from associated ribbon Schur functions. In addition, they

206 L.J. Billera et al. / Advances in Mathematics 204 (2006) 204–240

arise as sln-characters of the irreducible components of the Yangian representations inlevel 1 modules of sln [8].

In the theory of noncommutative symmetric functions of Gel’fand et al. [4], thenoncommutative analogues of the ribbon Schur functions form a homogeneous linearbasis. It is therefore of some interest to know what relations are introduced whenpassing to the commutative case. In particular, which pairs become identical? In [12,Exercise 7.56(b)] the ribbon Schur function indexed by a composition and its reversalare seen to be identical. However, as we shall see, this is not the whole story.

This paper is organized as follows. In Section 2, we introduce an equivalence relationon compositions and derive some of its properties. The relation is defined in terms ofcoefficients of symmetric functions when expressed in terms of the fundamental basisof the algebra of quasisymmetric functions. We show that this relation can be viewedcombinatorially in terms of coarsenings of the respective compositions. Theorem 2.6then shows compositions to be equivalent if and only if their corresponding ribbonSchur functions are identical.

Section 3 introduces a binary operation on compositions. In the case of compositionsdenoting the descent sets of a pair of permutations, the operation results in the com-position giving the descent set of their tensor product. In Sections 4 and 5 we proveour main result, Theorem 4.1, which states that equivalence of two compositions isprecisely given by reversal of some or all of the terms in some factorization. Thus thecongruence classes all have size given by a power of two; this power is the numberof nonsymmetric terms in the finest factorization of any composition in this class.

Finally, in Section 6, we consider the cone of F-positive symmetric functions, show-ing the Schur functions to be among its extremes and conjecturing its facets to be inone-to-one correspondence with equivalence classes of compositions.

The remainder of this section contains the basic definitions we will be using. Wherepossible, we are using the notation of [10] or [12].

1.1. Partitions and compositions

A composition � of n, denoted � � n, is a list of positive integers �1�2 . . . �k suchthat �1 + �2 + · · · + �k = n. We refer to each of the �i as components, and say that� has length l(�) = k and size |�| = n. If the components of � are weakly decreasingwe call � a partition, denoted � � n and refer to each of the �i as parts. For anycomposition � there will be two other closely related compositions that will be ofinterest to us. The first is the reversal of �, �∗ = �k . . . �2�1, and the second is thepartition determined by �, �(�), which is obtained by reordering the components of �in weakly decreasing order, e.g. �(3243) = 4332. Moreover we say two compositions�, � determine the same partition if �(�) = �(�).

Any composition � � n also naturally corresponds to a subset S(�) ⊆ [n − 1] ={1, 2, . . . , n − 1} where

S(�) = {�1, �1 + �2, �1 + �2 + �3, . . . , �1 + �2 + · · · + �k−1}.


Similarly any subset S = {i1, i2, . . . , ik−1} ⊆ [n − 1] corresponds to a composition�(S) � n where

�(S) = i1(i2 − i1)(i3 − i2) . . . (n − ik−1).

Finally, recall two partial orders that exist on compositions. For compositions �, � � n,we write � � � when � is lexicographically less than �, that is, � = �1�2 · · · �=�1�2 · · · = �, and the first i for which �i �= �i satisfies �i < �i . In particular,11 · · · 1 � � � n for any � � n. Secondly, given any two compositions � and � we say� is a coarsening of �, denoted ��, if we can obtain � by adding together adjacentcomponents of �, e.g., 3242�3212111. Equivalently, we can say � is a refinementof �.

1.2. Quasisymmetric and symmetric functions

We denote by Q the algebra of quasisymmetric functions over Q, that is all boundeddegree formal power series F in variables x1, x2, . . . such that for all k and i1 < i2 <

· · · < ik the coefficient of x�1i1

x�2i2

. . . x�k

ikis equal to that of x

�11 x

�22 . . . x

�k

k . There aretwo natural bases for Q both indexed by compositions � = �1�2 . . . �k , �i > 0: themonomial basis spanned by M0 = 1 and all power series M� where

M� =∑

i1<i2<···<ik

x�1i1

x�2i2

. . . x�k

ik

and the fundamental basis spanned by F0 = 1 and all power series F� where

F� =∑��

M�.

Note that Q is a graded algebra, with Qn = spanQ{M� | � � n}.We define the algebra of symmetric functions � to be the subalgebra of Q spanned

by the monomial symmetric functions

m� =∑

�:�(�)=�

M�, � � n, n > 0 (1.3)

and m0 = 1. Again, � is graded, with �n = � ∩ Qn.From quasisymmetric functions we can define Schur functions, which also form a

basis for the symmetric functions, but first we need to recall some facts about tableaux.For further background about tableaux, see [2].

For any partition � = �1 . . . �k � n the related Ferrers diagram (by abuse of notationalso referred to as �) is an array of left justified boxes with �1 boxes in the first row,


�2 boxes in the second row, and so on. For example, the Ferrers diagram 4332 is

A (Young) tableau of shape � and size n is a filling of the boxes of � with positiveintegers. If the rows weakly increase and the columns strictly increase we say it is asemi-standard tableau, and if in addition, the filling of the boxes involves the integers1, 2, . . . , n appearing once and only once we say it is a standard tableau. Note that inthis instance both the rows and columns strictly increase.

More generally, we can define skew diagrams and skew tableaux. Let �, � be parti-tions such that if there is a box in the (i, j)th position in the Ferrers diagram � thenthere is a box in the (i, j)th position in the Ferrers diagram �. The skew diagram �/�is the array of boxes {c | c ∈ �, c /∈ �}. For example, the skew diagram 4332/221 is

We can then define skew tableaux, semi-standard skew tableaux, and standard skewtableaux analogously.

Given a standard tableau or skew tableau T, we say it has a descent in position i ifi + 1 appears in a lower row than i. Denote the set of all descents of T by D(T ). Wetake [12, Theorem 7.19.7] as our definition of skew Schur functions.

Definition 1.1. Let �, � � n such that �/� is a skew diagram. Then the skew Schurfunction s�/� is defined by

s�/� =∑T

F�(D(T )), (1.4)

where the sum is over all standard tableaux T of shape �/�.

The Schur functions s� are those skew Schur functions with � = 0. For example,s22 = F22 + F121.

A skew diagram is said to be connected if, regarded as a union of squares, it has aconnected interior. If the skew diagram �/� is connected and contains no 2×2 array of


boxes we call it a ribbon. Observe ribbons of size n are in one-to-one correspondencewith compositions � of size n by setting �i equal to the number of boxes in the ith rowfrom the bottom. For example, the skew diagram 4332/221 is a ribbon, correspondingto the composition 2212.

Henceforth, we will denote ribbons by compositions, and denote the skew Schurfunctions s�/�, for a ribbon �/�, as the ribbon Schur function r�, where � is thecorresponding composition. Thus r2212 := s4332/221.

Further details on symmetric functions can be found in [12].

2. Equality of ribbon Schur functions

Although, in general, it is difficult to determine when two skew Schur functions areequal, it transpires that when computing ribbon Schur functions equality is determinedvia a straightforward equivalence on compositions.

2.1. Relations on ribbon Schur functions

There is a useful representation of ribbon Schur functions in terms of the basisof complete homogeneous symmetric functions h� = h�1h�2 · · · h�k

, known already toMacMahon [11, §168]. It can be derived from the Jacobi–Trudi identity [12, Theorem7.16.1]

s�/� = det(h�i−�j −i+j ), (2.1)

where h0 = 1 and hk = 0 if k < 0.

Proposition 2.1. For any � � n,

r� = (−1)l(�)∑��

(−1)l(�)h�(�).

Proof. Applying (2.1) to the ribbon shape �/� corresponding to � = �1 · · · �k , we get

r� = det

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

h�kh�k−1+�k

h�k−2+�k−1+�k· · · h�1+···+�k

1 h�k−1 h�k−2+�k−1 · · · h�1+···+�k−1

1 h�k−2 · · · h�1+···+�k−2

. . ....

1 h�1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.


Expanding down the first column gives

r� = r�1···�k−1h�k− r�1···�k−1+�k

,

which along with induction on l(�) gives the desired result. �

By inverting the relation in Proposition 2.1, we see immediately that the ribbonSchur functions r� generate �.

It is straightforward to establish all the algebraic relations that hold among ribbonSchur functions. Using Proposition 2.1, one can show that ribbon Schur functions satisfythe multiplicative relations

r� r� = r�·� + r��, (2.2)

where for � = �1 · · · �k and � = �1 · · · �l , � ·� = �1 . . . �k�1 . . . �l is the usual operationof concatenation and � � � = �1 . . . �k−1(�k + �1)�2 . . . �l , is near concatenation,which differs from concatenation in that the last component of � is added to the firstcomponent of �. Relation (2.2) has been known since MacMahon [11, §169]. Proofsof (2.2) in the noncommutative setting can be found in [4, Proposition 3.13] and[1, Proposition 4.1]. We show next that these relations generate all the relations amongribbon Schur functions.

Proposition 2.2. Let z�, � � n, n�1 be commuting indeterminates. Then as algebras,� is isomorphic to the quotient

Q[z�]/〈z�z� − z�·� − z��〉.

Proof. Consider the map � : Q[z�] → � defined by z� → r�. This map is surjectivesince the r� generate �. Grading Q[z�] by setting the degree of z� to be n = |�| makes� homogeneous. To see that � induces an isomorphism with the quotient, note thatQ[z�]/〈z�z� − z�·� − z��〉 maps onto Q[z�]/ ker � � �, since 〈z�z� − z�·� − z��〉 ⊂ker �.

It will suffice to show that the degree n component of Q[z�]/〈z�z� − z�·� − z��〉is generated by the images of the z�, � � n, and so has dimension at most the numberof partitions of n. We have the relations

z�·� + z�� = z�·� + z�� (2.3)

by commutativity of the z�. Let � = g1 . . . gk � n. We will show by induction on k thatz� = z�(�)+ a sum of z�’s with � having no more than k − 1 parts.


Let gi be a maximal component of �. Then by (2.2) and (2.3) we have

z� = zgi ...gkg1...gi−1 + a sum of z�’s with k − 1 or fewer parts

= zgizgi+1...gkg1...gi−1 + a sum of z�’s with k − 1 or fewer parts

= zgiz�(gi+1...gkg1...gi−1) + a sum of z�’s with k − 1 or fewer parts

= z�(�) + a sum of z�’s with k − 1 or fewer parts,

where the third equality uses the induction hypothesis. A trivial induction on the lengthof � now shows that any z�, � � n can be reduced in the quotient to a linear combinationof z�, � � n. �

As a consequence of the proof we get that the ribbon Schur functions r�, � � n, span�n and so form a basis. However, for general compositions � and �, it may be thatr� = r�. The rest of this section begins to deal with the question of when this canoccur. In principle, the relation r� = r� is a consequence of relations (2.2), and morespecifically (2.3), so a purely algebraic development of the main results of this papershould be possible. This has not yet been done.

2.2. Equivalence of compositions

We define an algebraic equivalence on compositions and reinterpret it in a combina-torial manner.

Definition 2.3. Let �, � be compositions. We say � and � are equivalent, denoted� ∼ �, if for all F = ∑

c�F� ∈ �, c� = c�.

That is, � ∼ � if F� has the same coefficient as F� in the expression of everysymmetric function. Note that any basis for � can be used as a finite test set for thisequivalence. We will be particularly interested in the monomial symmetric functionbasis (1.3) and the Schur function basis (1.4).

Example 2.1. For � = 211 and � = 121 we find that � �∼ � since s22 = F22 + F121.

For any composition � � n, we define M(�) to be the multiset of partitions deter-mined by all coarsenings of �, that is,

M(�) = {�(�) | ��}. (2.4)

We denote by multM(�)(�) the multiplicity of � in M(�).

Example 2.2. Note that while 2111 and 1211 have identical sets of partitions aris-ing from their coarsenings, M(2111) �= M(1211) since multM(2111)(311) = 1 whilemultM(1211)(311) = 2.


With this in mind we reformulate our equivalence. Recall that for F ∈ Q,

F =∑

c�F� =∑

d�M�,

where the c� and d� are related by

d� =∑��

c�, c� =∑��

(−1)l(�)−l(�)d�, (2.5)

and that F ∈ � if and only if d� = d� whenever �(�) = �(�). The following is a directconsequence of (1.3) and (2.5).

Proposition 2.4. If the monomial symmetric function m� = ∑�� n c�F�, then

c� = [m�]F� = (−1)l(�)−l(�) multM(�)(�),

that is, up to sign, c� is the multiplicity of � in the multiset M(�).

As an immediate consequence we get

Corollary 2.5. If � and � are compositions, then � ∼ � if and only if M(�) = M(�).

Example 2.3. Returning to the example � = 211 and � = 121, it is now straightforwardto deduce � �∼ � since

M(�) = {4, 31, 22, 211} �= {4, 31, 31, 211} = M(�).

We are now ready to state the main result of this section. Note first thatProposition 2.1 can be written

r� = (−1)l(�)∑

�∈M(�)

(−1)l(�)h�. (2.6)

Theorem 2.6. For the ribbon Schur functions r� and r� corresponding to compositions� and �, we have r� = r� if and only if M(�) = M(�).

Proof. If M(�) = M(�), then by (2.6), r� = r�. Conversely, since the hk arealgebraically independent, equality of r� and r� implies, again by (2.6), that M(�) =M(�). �

An immediate corollary of this and (1.1) are the following Littlewood–Richardsoncoefficient identities.


Corollary 2.7. Suppose ribbon skew shapes �/� and / correspond to compositions� and �, where � ∼ � are both compositions of n. Then, for all partitions � of n,

c��,� = c

,�.

Example 2.4. Since M(211) = {4, 31, 22, 211} = M(112) the above theorem assuresus that s222/11 = s4331/2221 and so, by Corollary 2.7, c222

11,� = c43312221,� for all partitions �

of 4.

An immediate consequence of Theorem 2.6, Corollary 2.5 and [12, Corollary 7.23.4]is another description of the equivalence ∼. For � = �(1)�(2) . . . �(n) ∈ Sn, the descentset of � is defined to be the set d(�) := {i | �(i) > �(i + 1)} ⊂ [n − 1].

Corollary 2.8. For �, � � n, � ∼ � if and only if for all � � n, the number of permu-tations � ∈ Sn satisfying d(�) = S(�) and d(�−1) = S(�) is equal to the number ofpermutations � ∈ Sn satisfying d(�) = S(�) and d(�−1) = S(�).

3. Compositions of compositions

In this section we describe a method to combine compositions into larger ones thatcorresponds to determining the descent set of the tensor product of two permutations.This leads naturally to a necessary and sufficient condition for two compositions to beequivalent.

3.1. Composition, tensor product, and plethysm

Let Cn denote the set of all compositions of n and let

C =⋃n�1

Cn.

Given � = �1 . . . �k � m and � = �1 . . . �l � n, recall the binary operations of concate-nation

· : Cm × Cn → Cm+n

(�, �) → � · � = �1 . . . �k�1 . . . �l

and near concatenation

� : Cm × Cn → Cm+n

(�, �) → � � � = �1 . . . �k−1(�k + �1)�2 . . . �l .


For convenience we write

��n = � � � � . . . � �︸︷︷︸n

.

These two operations can be combined to produce a third, which will be our focus:

◦ : Cm × Cn → Cmn

(�, �) → � ◦ � = ��1 · ��2 · · · ��k .

Example 3.1. If � = 12, � = 12 then � · � = 1212, � � � = 132 and � ◦ � = 12132.

It is straightforward to observe that C is closed under ◦ and that for � � m we have1 ◦ � = � ◦ 1 = �. Note that the operation ◦ is not commutative since 12 ◦ 3 = 36whereas 3 ◦ 12 = 1332.

We now see that composing compositions corresponds to determining descent setsin the tensor product of permutations.

Definition 3.1. Let � = �(1)�(2) . . . �(m) ∈ Sm and � = �(1)�(2) . . . �(n) ∈ Sn. Thentheir tensor product is the permutation

� ⊗ � = [(�(1) − 1)n + �(1)][(�(1) − 1)n + �(2)] . . . [(�(1) − 1)n + �(n)][(�(2) − 1)n + �(1)] . . . [(�(m) − 1)n + �(n)] ∈ Smn.

Remark 3.2. An alternative realization is as follows. Given � ∈ Sm, � ∈ Sn and them × n matrix

Mmn =

⎛⎜⎜⎜⎜⎜⎜⎝1 2 · · · n

n + 1 n + 2 · · · 2n

......

. . ....

(m − 1)n + 1 (m − 1)n + 2 · · · mn

⎞⎟⎟⎟⎟⎟⎟⎠then �Mmn is the matrix in which the ith row of �Mmn is the �(i)th row of Mmn.Similarly, M�

mn is the matrix in which the jth column of M�mn is the �(j)th column

of Mmn. With this in mind, � ⊗ � ∈ Smn is the permutation obtained by reading theentries of �M�

mn by row.


Example 3.3. If � = 213, � = 132 ∈ S3 then � ⊗ � = 213 ⊗ 132 = 465132798, and

Mmn =

⎛⎜⎜⎝1 2 3

4 5 6

7 8 9

⎞⎟⎟⎠ , �Mmn =

⎛⎜⎜⎝4 5 6

1 2 3

7 8 9

⎞⎟⎟⎠ , �M�mn =

⎛⎜⎜⎝4 6 5

1 3 2

7 9 8

⎞⎟⎟⎠ .

The following shows that the operation ◦ on compositions yields the descent set ofthe tensor product of two permutations from their respective descent sets.

Proposition 3.2. Let � ∈ Sm and � ∈ Sn. If d(�) = S(�) and d(�) = S(�) thend(� ⊗ �) = S(� ◦ �).

Proof. Let d(�) = S(�) = {i1, i2, . . . , ik} and d(�) = S(�) = {j1, j2, . . . , jl}. Thend(� ⊗ �) = {j1, j2, . . . jl, n + j1, n + j2, . . . , n + jl, . . . , (m − 1)n + j1, (m − 1)n +j2, . . . , (m − 1)n + jl} ∪ {ni1, ni2, . . . , nik} = S(� ◦ �). �

From Proposition 3.2 and the associativity of ⊗, we can conclude that ◦ is associative.Consequently we obtain

Proposition 3.3. (C, ◦) is a monoid.

Finally we relate the operation ◦ on compositions to the operation of plethysm onsymmetric functions. For the power sum symmetric function pm ∈ �m and g ∈ �n,define the plethysm

pm ◦ g = pm[g] = g(xm1 , xm

2 , . . .).

Extend this to define f ◦ g ∈ �mn for any f ∈ �m and g ∈ �n by requiring that themap taking f to f ◦ g be an algebra map. (See [10, p. 135; 12, p. 447] for details.)

When � � m and � � n the ribbon functions r� ◦ r� and r�◦� both have degree mn.While they are not in general equal, they are equal on the average, as seen by thefollowing identity.

Proposition 3.4. For any � � n,

∑��m

r� ◦ r� =∑��m

r�◦�.

Proof. We first note that

rm1 =

∑��m

r�,


which can be seen, for example, by repeated application of (2.2). Since the plethysmf ◦ g gives an algebra map in f, it follows that

∑��m

r� ◦ r� = rm1 ◦ r� = rm

� .

On the other hand, we show rm� = ∑

��m r�◦� by induction on m. The case m = 1 isclear. For m > 1, we get by induction and (2.2) that

rm� = r� ·

∑��m−1

r�◦�

=∑

��m−1

[r�·(�◦�) + r��(�◦�)

]=

∑��m

r�◦�. �

3.2. Unique factorization and other properties

If a composition � is written in the form �1 ◦ �2 ◦ · · · ◦ �k then we call this adecomposition or factorization of �. A factorization � = � ◦ � is called trivial if any ofthe following conditions are satisfied:

(1) one of �, � is the composition 1,(2) the compositions � and � both have length 1,(3) the compositions � and � both have all components equal to 1.

Definition 3.5. A factorization � = �1 ◦ · · · ◦ �k is called irreducible if no �i ◦ �i+1 isa trivial factorization, and each �i admits only trivial factorizations. In this case, each�i is called an irreducible factor.

Theorem 3.6. The irreducible factorization of any composition is unique.

Proof. We proceed by induction on the number of irreducible factors in a decomposi-tion.

First observe that if the only irreducible factor of a composition is itself then itsirreducible factorization is unique.

Now let � be some composition with two irreducible factorizations

�1 ◦ · · · ◦ �k−1 ◦ �k = � = �1 ◦ · · · ◦ �l−1 ◦ �l ,

and for convenience set � = �1 ◦ · · · ◦ �k−1, � = �k , � = �1 ◦ · · · ◦ �l−1 and = �l so

� ◦ � = � = � ◦ .


Our first task is to establish |�| = | | from which the induction will easily follow. Firstassume |�| = n and | | = s such that s �= n and without loss of generality let s < n.

If = s then it follows � �= n as if � = n then by our induction assumption and thefact that s �= n we have that the lowest common multiple of s and n would also be anirreducible factor, which is a contradiction. Hence � �= n and so l(�) > 1. Furthermoresince l(�) > 1 then � = �1 . . . �k must consist of components of � (the right-hand mostand k − 1 left-hand most components, for example), which implies s|�1, . . . , s|�k andhence � is not an irreducible factor.

Thus �= s so l( ) > 1 and since s < n we also have that l(�) > 1. In addition,since l(�) > 1, l( ) > 1 we have as above that � and must consist of components of�. Hence if s|n then it follows that � has as an irreducible factor and hence � is notan irreducible factor.

Consequently we have that if s �= n then l(�) > 1, l( ) > 1 and s�n. Moreover, thecomponents of � consist of the components of repeated (and perhaps the sum of thefirst and last components of ) plus one copy of truncated at one end of �. However,since � and consist of components of � it follows that if s �= n, l(�) > 1, l( ) > 1and s�n, then cannot be an irreducible factor. Thus |�| = n = s = | |.

Now that we have established |�| = | | we will show that in fact � = . If � = n

then clearly = n and we are done. If not, then since the last component of �, ��1it follows the right-hand components of � whose sum is less than n must be those of� and and since |�| = | | it follows that � = .

Since we now have � ◦ � = � = � ◦ �, it is straightforward to see � = �. By theassociativity of ◦ the result now follows by induction. �

We can also deduce expressions for the content and length of a composition interms of its decomposition. We omit the proofs, which each follow by a straightforwardinduction.

Proposition 3.7. For compositions �1, �2, . . . , �k

|�1 ◦ �2 . . . ◦ �k| =k∏

i=1

|�i |.

Proposition 3.8. For compositions �1, �2, . . . , �k

l(�1 ◦ �2 . . . ◦ �k) = l(�1) +k∑

i=2

⎛⎝i−1∏j=1

|�j |⎞⎠ (l(�i ) − 1).

Finally, it will be useful to observe that reversal of compositions commutes with thecomposition. The proof is clear.


Proposition 3.9. Let �, � be compositions then

(� ◦ �)∗ = �∗ ◦ �∗.

Remark 3.4. For � ∈ Sn, define �∗ ∈ Sn by �∗(i) := (n + 1) − �(n + 1 − i). It is easyto see that for � ∈ Sm, � ∈ Sn (� ⊗ �)∗ = �∗ ⊗ �∗ and �(d(�∗)) = (�(d(�)))∗. Onewonders whether this, in conjunction with Proposition 3.2, can provide a more directapproach to that of the next two sections.

4. Equivalence of compositions under ◦

We show in this section that the equivalence relation of Definition 2.3 is related tothe composition of compositions via reversal of terms. In particular, we prove

Theorem 4.1. Two compositions � and � satisfy � ∼ � if and only if for some k,

� = �1 ◦ �2 ◦ · · · ◦ �k and � = �1 ◦ �2 ◦ · · · ◦ �k,

where, for each i, either �i = �i or �i = �∗i . Thus the equivalence class of a composition

� will contain 2r elements, where r is the number of nonsymmetric (under reversal)irreducible factors in the irreducible factorization of �.

Before we embark on the proof, which will consist of the remainder of this sectionand the next, we note a corollary that follows immediately from Corollary 2.5, Theorem2.6, Corollary 2.8 and Theorem 4.1.

Corollary 4.2. The following are equivalent for a pair of compositions �, �:

(1) r� = r�,(2) in all symmetric functions F = ∑

c�F�, the coefficient of F� is equal to thecoefficient of F�,

(3) M(�) = M(�),(4) the number of permutations � ∈ Sn satisfying d(�) = S(�) and d(�−1) = S(�) is

equal to the number of permutations � ∈ Sn satisfying d(�) = S(�) and d(�−1) =S(�) for all �,

(5) for some k,

� = �1 ◦ �2 ◦ · · · ◦ �k and � = �1 ◦ �2 ◦ · · · ◦ �k,

and, for each i, either �i = �i or �i = �∗i .

Example 4.1. Since 12132 has irreducible factorization 12 ◦ 12, Corollary 4.2 assuresus that

r12132 = r13212 = r21231 = r23121


and, moreover, these are the only ribbon Schur functions equal to r12132. In addition,from

s54221/311 = r12132 = r13212 = s54431/332,

we can conclude from (1.1) the identity of Littlewood–Richardson coefficients

c54221311,� = c54431

332,�

for all partitions � of 9.

4.1. Reversal implies equivalence

We recall that for compositions � and �, � ∼ � if and only if M(�) = M(�) byCorollary 2.5. From this and Proposition 3.9 it is easy to conclude

Proposition 4.3. For compositions � and �1, . . . , �k ,

�∗ ∼ �

and

�1 ◦ �2 · · · ◦ �k ∼ �∗1 ◦ �∗

2 · · · ◦ �∗k.

We show now that reversal of any of the terms in a decomposition of � yields acomposition equivalent to �.

Theorem 4.4. For any compositions �, � and �,

(1) �∗ ◦ � ∼ � ◦ �,(2) � ◦ �∗ ∼ � ◦ � and(3) � ◦ �∗ ◦ � ∼ � ◦ � ◦ �.

Proof. By definition,

� ◦ � = ��1 · ��2 · · · ��k

and

�∗ ◦ � = ��k · · · ��2 · ��1 .

To prove (1), note that any coarsening � of � ◦ � that does not involve adding termsin different components ��i clearly corresponds to a coarsening of �∗ ◦ � that has the


same sorting �(�). On the other hand, a coarsening that involves, say, combining termsin ��i with terms of ��i+1 can be viewed as a coarsening of the first sort of

(�1, . . . , �i−1, �i + �i+1, �i+2, . . . , �k) ◦ �,

which can be seen to correspond to one arising as a coarsening of �∗ ◦ �.Assertions (2) and (3) follow from (1) and Proposition 4.3 via

� ◦ �∗ ∼ �∗ ◦ � ∼ � ◦ �

and

� ◦ �∗ ◦ � ∼ �∗ ◦ � ◦ �∗ ∼ � ◦ �∗ ◦ �∗ ∼ � ◦ � ◦ �,

respectively. �

One direction in the assertion of Theorem 4.1 now follows from Theorem 4.4. Theremainder of this section and the next is devoted to the proof of the other direction.

4.2. Equivalence implies reversal

In this subsection, we prove the converse to the result established in the previoussubsection: namely, that if � ∼ �, then there is a factorization � = �1 ◦ · · · ◦ �k suchthat � = �1 ◦ · · · ◦ �k , where �i = �i or �∗

i . We achieve this via two theorems. The firstof these is

Theorem 4.5. Let � ∼ �, and � = �◦ . Then � can be decomposed as �◦ with � ∼ �and ∼ .

Example 4.2. Let � = 13212 and � = 12132. It is straightforward to check that thesetwo compositions are equivalent. Note we have that � = 21 ◦ 12. Theorem 4.5 saysthat there should be a decomposition � = � ◦ with � ∼ 21 and ∼ 12. We observethat � = 12 ◦ 12 satisfies these conditions.

In order to prove Theorem 4.5 we require two lemmas:

Lemma 4.6. Let � = � ◦ where � � n. Let have size m and p components. Let� = �1 . . . �k be a partition of n which occurs in M(�). Let �i be the remainder when�i is divided by m, and suppose that the sum of the �i is m. Then the number ofnonzero �i is at most p.

Proof. Reordering the parts of � if necessary, let �1 . . . �k be a composition of n whichis a coarsening of �. Now consider the composition of m given by �1 . . . �k (where


we omit any zero components). This composition is a coarsening of , and thus has atmost p components. �

Lemma 4.7. Let � = � ◦ where � � n and � m, and let � be a partition of m, withk parts. Then

multM(�)(�, n − m) = (k − 1 + multM(�)(m, n − m))multM( )(�).

Remark 4.3. Note that in the statement of the previous lemma and subsequently, whenthe context is unambiguous, we will refer to the multiplicity of a composition in themultiset of coarsenings of a composition when we intend the multiplicity of the partitiondetermined by that composition.

Proof of Lemma 4.7. Given a way to realize � from , there are k−1+multM(�)(m, n−m) corresponding ways to realize (�, n − m) from �: one must pick where to put inthe n − m component. �

Proof of Theorem 4.5. Let the size of be m. Write q = n/m. Let the number ofcomponents of be p.

Define � � q by setting S(�) = {i | mi ∈ S(�)}. Now multM(�)(�) = multM(�)m�where we write m� for the partition obtained by multiplying all the parts of � by m.Similarly, multM(�)(�) = multM(�)m�. Thus, the equivalence of � and � follows fromthat of � and �.

Define i � m, i = 0, . . . , q − 1, by setting

S(i ) = {x | 0 < x < m, x + im ∈ S(�)}.

We wish to show that all the i are equal and equivalent to .For any 0� i�q − 1, the number of components of i is at most p: otherwise,

consider the composition of � consisting of

• im plus the first component of i ,• the remaining components of i except the last,• the last component of i plus (q − 1 − i)m.

The partition corresponding to this composition appears in M(�) but by Lemma 4.6,it cannot appear in M(�), which is a contradiction.

The cardinalities of S(�) and S(�) must be the same, and we have already seen that|S(�) ∩ mZ| = |S(�) ∩ mZ|. We know that |S(�) ∩ (Z \ mZ)| = q(p − 1), so the samemust hold for �. Now, since each of the i has at most p components, each of the i

must have exactly p components.


We now need the following lemma:

Lemma 4.8. Let �, �, and the i be as already defined. Let 0� i < j �q − 1. LetS(i ) = {a1 < · · · < ap−1} and S(j ) = {b1 < · · · < bp−1}. Then at �bt for all t.

Proof. If this were not so, let � be the partition consisting of the following:

• im plus the first component of i ,• the second through tth components of i ,• (j − i)m + bt − at ,• the t + 1th through (p − 1)th components of j ,• the last component of j plus (q − j − 1)m.

Now � appears in M(�) but by Lemma 4.6 does not appear in M(�), a contradiction.�

We now return to the proof of Theorem 4.5. Let � be the partition of m determinedby . Let x = multM(�)(m, n − m) ∈ {0, 1, 2}. The multiplicity of (�, n − m) in M(�)

is p − 1 + x.Now consider the possible occurrences of (�, n − m) in M(�). If the tth element of

S(0) coincides with the tth element of S(q−1), then we have one possible occurrenceof (�, n−m) with n−m as the (t + 1)th component. Also, since by the equivalence of� and �, x of {m, n − m} are in S(�), there are x possible occurrences of compositionsrealizing (�, n − m) such that the n − m part is either the first or the last component.However, there must be p − 1 + x realizations of (�, n − m), so all these possibilitiesmust actually realize the partition.

In particular, this shows that S(0) and S(q−1) must coincide. Now, by Lemma 4.8,all the S(i ) must coincide, and we can now denote all the i by . The equal-ity of the i (in particular, the equality of 0 and q−1) means that we can applythe same argument as in Lemma 4.7 to show that for � a partition of m with kparts,

multM(�)(�, n − m) = (k − 1 + x)multM()(�).

The equivalence of � and � also implies the multiplicities of � in M() and M( )are equal for any � that is a partition of m, and hence that and are equivalent, asdesired. This completes the proof of the theorem. �

The second theorem requires the concept of reconstructibility of a composition.

Definition 4.9. A composition � is said to be reconstructible if knowing M(�) allowsus to determine � up to reversal.

Example 4.4. The composition 112 is reconstructible because if � is a compositionsatisfying multM(�)�(211) = 1 and multM(�)�(22) = 1, then � = 112 or � = 211.


Theorem 4.10. If � � n is not reconstructible, then � decomposes as �◦ , where neither� nor have size 1.

Proof. We establish this result by defining a function h on � and then proving thatif � is not reconstructible then h is periodic with period | | > 1. This, in turn, yieldsour result. Since the proof of the periodicity of h is somewhat technical we will statethe pertinent lemmas but postpone their proofs until Section 5. Before we define h weneed a few other definitions.

Definition 4.11. With respect to a composition � � n, for any 0 < i < n, we say thati is of type 0, 1, or 2, depending on whether there are 0, 1, or 2 occurrences of thepartition (i, n − i) in M(�) or, equivalently, if 0, 1, or 2 of i, n − i are in S(�). Fori = n/2, if n/2 is an integer, we say that its type is twice the number of occurrencesof (n/2, n/2) in M(�).

Example 4.5. In the composition 11231, 1, 4 and 7 are type 2, 2 and 6 are type 1, 3and 5 are type 0.

Fix a composition � of n. Let Ai be the set of those elements of [n − 1] that areof type i with respect to �. If A1 = ∅, then clearly � is reconstructible. Note thatA1 = ∅ exactly when � is symmetric under reversal. Now suppose A1 �= ∅. Let k bethe least element of A1. Reversing � if necessary, we may assume that k ∈ S(�), andn − k /∈ S(�).

Definition 4.12. For j ∈ A1, we say that j is determined if we can tell whether or notj ∈ S(�) from M(�) and the knowledge that k ∈ S(�).

Example 4.6. In the composition 12132, A1 = [8]. The determined elements are{1, 2, 4, 5, 7, 8}. 3 and 6 are undetermined, because 12132 ∼ 13212, and 3 ∈ S(12132),while 3 /∈ S(13212), and the reverse is true of 6.

Definition 4.13. For x, y ∈ [n − 1], we say that they agree if they are of the sametype and either both or neither are in S(�).

Remark 4.7. Note that this second condition follows from the first for x, y of eventype.

We extend the notion of type to all Z by saying that multiples of n are type 0, andotherwise, x has the same type as x mod n.

If every element of A1 is determined, then � is reconstructible. Suppose � is notreconstructible, so there are undetermined elements of A1. Let us define T0 to be theset of all integers that are undetermined, where we extend the notion of determinednessto all integers by saying that, in general, x is determined if and only if x mod n isdetermined. Let t0 be the greatest common divisor of T0. We are going to defineinductively a collection Ti of sets of integers. We will write T� j for the union ofT0, . . . , Tj . Let tj be the greatest common divisor of T� j .


Definition 4.14. For i > 0, let Ti be the set of x not divisible by ti−1, such that thereis some t ∈ Ti−1 with x and x + t of even type and disagreeing.

Clearly, only finitely many of the Ti are nonempty. Let s be the greatest commondivisor of all the Ti . By convention, set t−1 = n.

We are now ready to define the function h and state the results needed in order toanalyze its periodicity. Let g and h be the functions defined on Z with respect to � by

h(x) =

⎧⎪⎪⎨⎪⎪⎩0 if x is type 0,

1 if x is type 1 and x mod n ∈ S(�),

−1 if x is type 1 and x mod n /∈ S(�),

2 if x is type 2

and

g(x) =⎧⎨⎩

0 if x is of even type,1 if x is type 1 and x mod n ∈ S(�),

−1 if x is type 1 and x mod n /∈ S(�).

Consider the following three statements concerning the functions g and h and thesets Ti .

Pi : The function g is ti-periodic except at multiples of ti .Qi : The function h is ti−1-periodic except at multiples of ti .Ri : For x ∈ Ti+1 and z of type 1, ti �z, z and x + z agree.

These statements are all defined for i�0. Note that Q0 is immediate, by our conven-tional definition of t−1. The remaining statements will follow by simultaneous induction.

Example 4.8. Consider the composition � = 132121332 = 213 ◦ 12. We can writeout the values of g and h on [18] as strings of 18 characters, writing + for 1, and− for −1.

g = + − 0 + − + + − 0 + − − + − 0 + −0

h = + − 0 + − + + − 2 + − − + − 0 + −0

Since � ∼ 312 ◦ 12, 6 and 12 are type 1 undetermined. In fact, T0 ∩ [18] = {6, 12};t0 = 6. We next observe that 3 and 9 belong to T1 because 3 and 3 + 6 (resp. 9 and9 + 6) are of even type but disagree, and 6 ∈ T0. In fact, T1 ∩ [18] = {3, 9}. Hencet1 = 3. All the Ti for i > 1 are empty. Thus ti = 3 for i > 1.

We now take a look at the meanings of Pi and Qi for this choice of �. P0 says thatg is 6-periodic except at multiples of 6. Q0 says h is 18-periodic except at multiplesof 6. P1 says that g is 3-periodic except at multiples of 3. Q1 says that h is 6-periodic


except at multiples of 3. P2 says nothing more than P1. Q2 says that h is 3-periodicexcept at multiples of 3.

For clarity of exposition, we will divide the proof of the simultaneous induction intoseveral parts:

• Proof of P0 (Lemma 5.7).• Proof that Pj and Qj for j � i imply Ri (Lemma 5.10).• Proof that Ri and Pi imply Pi+1 (Lemma 5.17).• Proof that Pi+1 and Qi imply Qi+1 (Lemma 5.18).

These four lemmas establish the simultaneous induction.Observe that for i sufficiently large, ti = ti−1 = s. Thus Qi implies that h is s-periodic

except at multiples of s. We now apply the following lemma:

Lemma 4.15. The composition � � n has a decomposition � = �◦ with | | = p if andonly if p divides n and the function h determined by � is p-periodic except at multiplesof p.

Proof. Suppose � has such a decomposition. It is clear that p|n. Write h� for thefunction determined by �, and h for the function determined by . For x ∈ [n − 1],p�x, h�(x) = h (x mod p), which proves the desired periodicity.

Conversely, suppose that h� has the desired periodicity. Define � p by settingh |[0,p−1] = h�|[0,p−1]. Define � � n/p by setting h�(x) = h�(px). It is then clearthat � = � ◦ . �

Example 4.9. Continuing Example 4.8, and applying Lemma 4.15 to the assertion ofQ2, that h is 3-periodic except at multiples of 3, we conclude that 132121332 = � ◦ where | | = 3, which is indeed true, since 132121332 = 213 ◦ 12.

Returning to the proof of Theorem 4.10, we see that an application of Lemma 4.15implies that � = � ◦ , where | | = s. We have s < n since � is not reconstructible.Since also s > 1 (see Lemma 5.20), this factorization is nontrivial. This provesTheorem 4.10. �

We are now in a position to prove our main result.

Proof of Theorem 4.1. If � and � satisfy � ∼ � then by Theorem 4.10, we can factor� = �1 ◦ · · · ◦�k where all the �i are reconstructible. Applying Theorem 4.5 repeatedly,we find that � = �1 ◦ · · · ◦ �k , where �i ∼ �i . However, since the �i are reconstructible,�i ∼ �i implies that �i = �i or �∗

i .Conversely, if � = �1 ◦ · · · ◦ �k and � = �1 ◦ · · · ◦ �k such that either �i = �i or

�i = �∗i then by Theorem 4.4 it follows that � ∼ �.

Finally, observe that by Theorem 3.6 the equivalence class of � contains 2r elementswhere r is the number of nonsymmetric compositions under reversal in the irreduciblefactorization of �. �


5. Technical lemmas

In this section we prove the technical lemmas which we deferred from the previoussection. We begin with a basic lemma which will be useful throughout this section.

Lemma 5.1. Let � � n, and let � = m ◦ � for some m > 1. Then:

(1) M(�) can be determined from M(�).(2) If n does not divide x, then x has the same type with respect to � as x mod n does

with respect to �.(3) The functions g and h determined by � and � coincide.(4) ti (�) = ti (�).(5) Each of Pi , Qi and Ri holds for � if and only if it holds for �.

Proof. Suppose we know M(�). We wish to determine M(�). This is equivalent todetermining the equivalence class of � with respect to equivalence for compositions. ByTheorem 4.5, the equivalence class of � consists exactly of those compositions whichcan be written as m ◦ � with � ∼ �. Thus, knowing M(�) suffices to determine M(�).

Observe that (2), (3), and (5) are immediate from the definitions. For (4), we have toverify that x is determined for � if and only if x mod n is determined for �. Supposex mod n is determined for �. That says exactly that all compositions in the equivalenceclass of � agree at x mod n. By Theorem 4.5, the equivalence class of � consists ofthe single-part partition m composed with elements of the equivalence class of �, andtherefore x is determined for �. The converse follows the same way. �

The purpose of this lemma is that at any step in the simultaneous induction thatproves Pi , Qi and Ri , we can replace � by m ◦ � if we so desire.

5.1. Proof of P0

In this subsection we prove P0 (Lemma 5.7). We also prove Lemma 5.8, which willbe necessary for our proof of Lemma 5.20.

Let the elements of T0 ∩ [n − 1] be m1 < · · · < ml . Let ri = gcd(m1, . . . , mi). Notethat m1 and n − m1 are both in T0, so rl divides n, and therefore rl coincides with t0,the greatest common divisor of T0. We begin with some lemmas.

Lemma 5.2. Suppose that x, y, and x + y all lie in A1. Then from M(�) we can tellif x, y, and n − (x + y) all agree, or if they don’t all agree.

Proof. If x, y, and n− (x +y) agree, then (x, y, n− (x +y)) does not appear in M(�).Otherwise, it does appear. �

Lemma 5.3. Suppose x, y, and x + y lie in [n − 1] and exactly two of them lie in A1.Then we can determine from M(�) whether or not they agree.

Proof. The proof is similar to that of Lemma 5.2, though there are more cases to check.It is sufficient to check the cases: x type 0 (and the others type 1); x type 2; x + y


type 0; x + y type 2. In each case, one sees that the multiplicity of (x, y, n − (x + y))

in M(�) depends on whether the two type 1 points agree or disagree. �

Definition 5.4. We say that a function f defined on a set of integers including [p − 1]is antisymmetric on [p − 1] if f (x) = −f (p − x) for 0 < x < p.

Lemma 5.5. Let f be a function on [d−1] which takes values 0, 1, −1, ∗, and supposethat there is some c < d , such that for all x for which both sides are well-defined

f (x) = −f (d − x), (5.1)

f (x) = −f (c − x), (5.2)

f (x) = f (c + x) (5.3)

except that if either side equals ∗, the equation is not required to hold. Further, werequire that the points where f takes the value ∗ are either exactly the multiples of cless than d, or else no points at all. Let r be the greatest common divisor of c and d.Then on multiples of r, f takes on only the values 0 and (possibly) ∗. On nonmultiplesof r, f is r-periodic, and f is antisymmetric on [r − 1].

Proof. The proof is by induction. We first consider the base case, which is when r = c.Periodicity is (5.3). Antisymmetry is (5.2). Notice (5.3) also implies that f is constanton multiples of c; by (5.1) this constant value is either ∗ or 0.

Now we prove the induction step. Let (5.1′), (5.2′), (5.3′) denote (5.1), (5.2), and(5.3), with d replaced by c and c replaced by d mod c. It is easy to see that (5.1′),(5.2′), and (5.3′) follow from (5.1), (5.2), (5.3). Also, f restricted to [c − 1] never takeson the value ∗. The desired results now follow by induction. �

Lemma 5.6. For 1� i� l,(i) g is antisymmetric on [ri − 1],(ii) g is ri-periodic on [mi − 1] except at multiples of ri .

Proof. The proof is by induction on i. We begin by proving the base case, which iswhen i = 1.

Suppose 0 < x < m1. By assumption, x and m1−x are determined if they are type 1.Suppose one of them is of even type, and the other is type 1. Then by Lemma 5.3, wecan determine m1, contradiction. Suppose that x and m1 − x are both type 1. If theyagree, Lemma 5.2 allows us to determine m1, contradiction. Hence they must disagree.This establishes (i) in the base case. In the base case, (ii) is vacuous.

Now we prove the induction step. For i�2 define a function gi on [mi − 1], asfollows:

gi(x) ={ ∗ if ri−1|x,

g(x) otherwise.


We wish to apply Lemma 5.5 to gi , with d = mi , c = ri−1. If 0 < x < mi , and neitherx nor mi − x is a multiple of ri−1 (so in particular, neither is type 1 undetermined),then, as in the proof of the base case, gi(x) = −gi(mi − x). This is condition (5.1).

Suppose both x and mi−1 +x < n are type 1 and determined. If they disagree (whichmeans that x and n − (mi−1 + x) agree), then we can determine mi−1, contradiction.Similarly, if one is of even type and the other is type 1 determined, we can determinemi−1, again a contradiction. It follows that gi is mi−1-periodic on [mi − 1] except atmultiples of ri−1. However, by induction, gi is ri−1-periodic except at multiples ofri−1 on [mi−1 − 1], so gi is ri−1-periodic except at multiples of ri−1 on [mi − 1]. Thisestablishes condition (5.3). Condition (5.2) follows by the induction hypothesis.

Thus, we can apply Lemma 5.5. This proves the induction step, and hence thelemma. �

Lemma 5.7. P0 holds, that is to say, g is t0-periodic except possibly at multiplesof t0.

Proof. Since t0 = rl , we have already shown (Lemma 5.6) that g is t0-periodic on[ml −1] except at multiples of t0. Since g is antisymmetric on [n−1] (by the definitionof g) it follows that g is t0-periodic on [n − 1] except at multiples of t0, from whichthe desired result follows. �

We now prove Lemma 5.8 which will be used in the proof of Lemma 5.20.

Lemma 5.8. The greatest common divisor t0 of T0 does not divide k.

Proof. Suppose otherwise. Let i be the least index such that ri |k. Note that i > 1,since k < m1. By the result of applying Lemma 5.5 to gi , we know that gi is zero onmultiples of ri which are not multiples of ri−1. However, this means that gi(k) = 0,which contradicts the fact that k is type 1. �

5.2. Proof of Ri

We begin by deducing R0 from P0 (Lemma 5.9). We then prove the general statementthat Pj and Qj for j � i imply Ri (Lemma 5.10), which reduces to the argument forLemma 5.9.

Lemma 5.9. P0 implies R0.

Proof. We must show that if z is type 1 and not a multiple of t0 (which means inparticular that it is determined), and y1 ∈ T1, then z and z + y1 agree.

Since y1 ∈ T1, there is some y0 ∈ T0 such that y1 and y1 + y0 are of even type anddisagree. Clearly, we may assume that z, y0, and y1 are all positive.

If z + y1 + y0 > n, we may replace � by m ◦ � for some sufficiently large m, byLemma 5.1. We also wish to assume that z < y0. If this is not true, we can make ittrue by another replacement as above, followed by adding n to y0.


By P0, we know that z and z + y0 agree. Also, observe that since z is type 1, sois n − z, and thus, by P0, so is any w ≡ −z mod t0. Since y1 is of even type, thismeans that t0�z + y1, so z + y1 is of even type or determined, and P0 tells us thatg(z + y1 + y0) = g(z + y1).

By considering the multiplicity of (y0, y1, z, n− (y1 +y0 + z)) in M(�), we see thatone of two things happens:

• z + y1 and z + y1 + y0 are type 1 and both agree with z• z + y1 agrees with y1, while z + y1 + y0 agrees with y0 + y1.

We now exclude the second possibility. Suppose we are in that case. Let w = y0−z. Thisw is not a multiple of t0, so w is of even type or is determined. As already remarked,since w ≡ −z mod t0, w must be type 1 determined. Now apply the previous part ofthe proof to (y′

0, y′1, z

′) with y′0 = y0, y′

1 = z + y1, z′ = w. Then we see that eitherw+z+y1 must either be the same type as y1, or as w. However, w+z+y1 = y0 +y1,and we know that it is of even type but disagrees with y1, which is a contradiction.

�

Lemma 5.10. Pj and Qj for j � i imply Ri .

Proof. We wish to show that for z of type 1, ti �z (so in particular z is determined),and x ∈ Ti+1, that z and z + x agree. Write yi+1 for x. Now there is some yi ∈ Ti

such that yi+1 and yi + yi+1 are of even type and disagree. Similarly, choose yj forall 0�j � i − 1 so that yj+1 and yj+1 + yj are even type and disagree.

For I a subset of [0, i + 1], write yI for the sum of the yj with j ∈ I . We nowdetermine the types of yI and yI + z.

Lemma 5.11. Let I ⊂ [0, i + 1]. Let j be the maximal element of I. Then:

(1) If I does not contain j − 1, then yI agrees with yj .(2) If I does contain j − 1, then yI is of even type disagreeing with yj .(3) Either z + yI is of even type or it is determined.(4) If I does not contain i + 1, z + yI agrees with z.(5) If I contains i + 1 but not i, z + yI agrees with z + yi+1.(6) If I contains i + 1 and i, z + yI agrees with z + yi+1 + yi .(7) Either z + yi+1 and z + yi+1 + yi agree, or they are both of even type.

Proof. Statement (1) follows from Qj−1, since yj is not a multiple of tj−1. Statement(2) follows because yj and yj + yj−1 are of even type and disagree, and then byapplying Qj−1 as before.

Since g is ti-periodic except at multiples of ti , and its period is anti-symmetric, itfollows that any w ≡ −z mod ti must be of odd type. Thus yi+1 �≡ −z mod ti . All theother yl are multiples of ti . Thus ti �z + yI , so z + yI is either determined or of eventype. This establishes (3).


Statement (4) follows from Pi , since z is not a multiple of ti . Since ti �z + yi+1,(5) follows from Qi . Statement (6) follows from Qi together with the fact that, sinceti does not divide z + yi+1, it does not divide z + yi+1 + yi . Statement (7) followsfrom Pi . �

We now return to the proof of Ri . We want to assume that p = n−(z+∑i+1j=0 yj ) > 0,

and that p does not coincide with any yj or z. In order to guarantee this, by Lemma 5.1,we may replace � by m ◦ �, and add multiples of n as desired to the yi and z.

Since y0 ∈ T0, it is undetermined. This means precisely that there is some compo-sition � which is equivalent to � (but not equal to �), such that k ∈ S(�), but y0 is inexactly one of S(�), S(�). Note that since � is equivalent to �, every 0 < x < n hasthe same type in � and �.

Write � for the partition of n whose parts are (y0, y1, . . . , yi+1, z, p). One con-sequence of the equivalence of � and � that we shall focus on is the fact thatmultM(�)(�) = multM(�)(�).

Let � be the set of all the compositions of n determining the partition �. It willbe convenient for us to keep track of such a composition as two lists: the left list,which consists of the components in order which precede p, and the right list, whichconsists of the components following p in reverse order. For any composition in �,each component other than p occurs in exactly one list, and any pair of lists with thisproperty determines a composition.

We put an order ≺ on the components yj , z by ordering the yj by their indices,and setting yj ≺ z for j �= i + 1. (Thus, the order is nearly a total order but not quite:yi+1 and z are incomparable.)

Definition 5.12. A composition in � is called ordered if both its right and left lists arein (a linear extension of) ≺ order. The other compositions in � are called disordered.

Lemma 5.13. The number of disordered compositions which can be obtained as coars-enings of � is the same as the number that can be obtained as coarsenings of �.

Proof. To prove this lemma, we will define an involution i on disordered compositionssuch that � is a coarsening of � if and only if i(�) is a coarsening of �.

Fix a disordered composition �. Let M(�) be the maximal subset of y0, . . . , yi+1, z

which is a ≺ order ideal such that M(�) consists of the union of initial subsequencesof the left and right lists of �, and these subsequences are in ≺ order. Write ML(�) andMR(�) for these two initial subsequences. Then i(�) is obtained by swapping ML(�)

and MR(�). Observe that i(�) is disordered if and only if � is disordered.

Example 5.1. We give an example of the definition of i.

If � =y0 y1y2 y4y5 y3z

then M(�) = {y0, y1, y2} and i(�) =y1 y0y5 y2z y4

y3

.


We shall now define a bijection, also denoted i, taking S(�) to S(i(�)), such thatfor x ∈ S(�), x ∈ S(�) if and only if i(x) ∈ S(�). The existence of such a bijectionbetween S(�) and S(i(�)) implies that � is a coarsening of � if and only if i(�) is acoarsening of �, proving the lemma.

To define the bijection between S(�) and S(i(�)), we need another definition:

Definition 5.14. We say x ∈ S(�) is an outside break if it is either the sum of aninitial subsequence of ML(�) or n minus the sum of an initial subsequence of MR(�).Otherwise, we say that x ∈ S(�) is an inside break.

Example 5.2. In our continuing example, the outside breaks of � are y0, y0 + y2, andn − y1, while the outside breaks of i(�) are n − y0, n − (y0 + y2), and y1. The insidebreaks in � are y0 + y2 + y5, y0 + y2 + y5 + z, n − (y1 + y4), n − (y1 + y4 + y3), whilethe corresponding inside breaks in i(�) are y1 + y5, y1 + y5 + z, n − (y0 + y2 + y4),n − (y0 + y2 + y4 + y3).

If x is an outside break of �, set i(x) = n − x. Clearly, i(x) is an outside breakof i(�). Now observe that all the outside breaks except y0 or n − y0 are of eventype in � by Lemma 5.11. Thus for these outside breaks (excluding y0 and n − y0),x ∈ S(�) if and only if x is type 2 for � if and only if x is type 2 for � if and onlyif i(x) is type 2 for � if and only if i(x) ∈ S(�). On the other hand, y0 ∈ S(�) ifand only if n − y0 ∈ S(�). Thus, for x an outside break of �, x ∈ S(�) if and only ifi(x) ∈ S(�).

Now we consider the inside breaks. Let yL denote the sum of the yj appearing inML(�), and similarly for yR . If x is an inside break for �, set i(x) = x − yL + yR .This is clearly an inside break for i(�).

Since � is disordered, define l by M(�) = {y0, y1, . . . , yl}. By definition, all the yj

that occur in yL and yR have j � l. To show that x ∈ S(�) if and only if i(x) ∈ S(�)there are four cases to consider: when x is of the form yI , z+yI , n−yI , or n−yI −z.In the first case, observe that I contains at least one element greater than l + 1, andso, by Lemma 5.11(1) or (2), x and i(x) agree and are of even type. It follows thatx ∈ S(�) if and only if i(x) ∈ S(�) if and only if i(x) ∈ S(�), as desired.

In the second case, since l� i − 1 it is again clear by Lemma 5.11(4), (5), or (6),that x and i(x) agree, so x ∈ S(�) if and only if i(x) ∈ S(�). By Lemma 5.11(3), i(x)

is either determined or of even type, so i(x) ∈ S(�) if and only if i(x) ∈ S(�), whichestablishes the desired result.

The third and fourth cases are similar to the first and second cases. This completesthe proof that i is a bijection from S(�) to S(�), which completes the proof of thelemma. �

Now we consider the ordered compositions. Suppose � is an ordered compositionwhich is a coarsening of �. Thus y0 is the beginning of one list. Which list is determinedby which of y0 and n − y0 is a break in �. Since y1 and y1 + y0 disagree, which listy1 occurs in is forced. Similarly for y2, etc. Hence all the yj are forced up to yi .There are now six possible ways to complete the construction. For each of these six


possibilities we show the positions of yi , yi+1, and z in the two lists.

(a) (b) (c) (d) (e) (f )yi yi z yi yi z yi yi+1 yi yi+1z yi+1 yi+1 yi+1 z z

yi+1 z

The argument now proceeds as in Lemma 5.9. Essentially what has happened is thatby reducing to ordered compositions, we do not need to consider the yj with j < i.We are now only interested in the middle part of the composition, which involvesparts yi , yi+1, z, and p. Also yi now behaves like y0 in Lemma 5.9: we count up thenumber of compositions which occur with yi on the extreme left (among the four partswe are interested in) and those where it occurs on the extreme right. One of thesenumbers represents the contribution of ordered partitions to multM(�)(�), the other thecontribution to multM(�)(�). These numbers must therefore be the same. As in theproof of Lemma 5.9, we consider cases based on the types (and for type 1, whether ornot each is a break) of yi+1, z, yi+1 +z, and yi+1 +yi +z. Lemma 5.11(7) eliminates anumber of possibilities and with the remainder, as in Lemma 5.9, one of the followingtwo things must happen:

• z + yi+1 and z + yi+1 + yi are type 1 and both agree with z,• z + yi+1 agrees with yi+1, while z + yi+1 + yi agrees with yi+1 + yi .

We now exclude the second possibility.Since z is not a multiple of ti , Pi tells us that yi − z agrees with n − z, which is

type 1 determined. Also by Pi , z �≡ −yi+1 (mod ti ), so ti does not divide z + yi+1.Since yi ∈ Ti , and z+yi+1 and z+yi+1 +yi disagree, z+yi+1 ∈ Ti+1. Set z′ = yi − z,y′i+1 = z + yi+1, y′

j = yj for j � i. Applying the whole proof of the lemma sofar, we find that z′ + y′

i+1 must agree either with z′ or y′i+1, which is to say that

yi + yi+1 agrees with either yi − z or z + yi+1, both of which are impossible, and weare done. �

5.3. Proofs of Pi and Qi

We begin with a preliminary lemma which will be useful for the proofs of Lem-mas 5.17 and 5.18. While working towards proving these two lemmas, we will oftenneed to consider Z/tj Z (for some j). We will write Ztj for Z/tj Z, and z for the imageof z in Ztj .

Lemma 5.15. Let f be a function defined on Z. Let S ⊂ Ztj be such that f (z) =f (z + tj−1) for z ∈ S. Suppose further that for any x ∈ Tj , f (z) = f (z + x) providedz ∈ S. Then f (z) = f (z + tj ) for all z ∈ S.

Proof. Write tj as the sum of a series of elements of T� j . Let the partial sums of thisseries be x1, x2, . . . , xm = tj . Then observe that if z ∈ S, then the same is true for z + xl


for all l. It follows from the assumptions of the lemma that f (z + xl) = f (z + xl+1),and the result is proven. �

Lemma 5.16. For p > 0, if x ∈ Tp, then there is an element x′ ∈ Tp such thatx ≡ −x′ modulo tp−1.

Proof. Since x ∈ Tp, there is some y ∈ Tp−1 such that x and y + x are of even typeand disagree. It follows that n − y − x and n − x are of even type and disagree, andhence that n − y − x ∈ Tp. Set x′ = n − y − x. �

Lemma 5.17. Ri and Pi imply Pi+1.

Proof. Let j = i + 1. We wish to show that g is tj -periodic except at multiples oftj . Let S = Ztj \ {0}. Pi tells us that g is tj−1-periodic except at multiples of tj−1.Suppose z ∈ S, and x ∈ Tj . Ri tells us that if z is type 1, then g(z + x) = g(z).Likewise, if z + x is type 1, then, choosing x′ as provided by Lemma 5.16, z + x + x′is type 1, and now by the tj−1 periodicity of g, g(z+ x) = g(z). If neither z nor z+ x

is type 1, then g(z + x) = 0 = g(z).Thus, it follows that for any z such that z ∈ S, and x in Tj , that g(z + x) = g(z).

Therefore, we can apply Lemma 5.15, and desired result follows. �

Lemma 5.18. Pi+1 and Qi imply Qi+1.

Proof. Let j = i. Let S = Ztj \ tj+1Ztj . We wish to show that h(z) = h(z + tj ) forz ∈ S. Qi tells us that h(z + tj−1) = h(z) for z ∈ S. Now suppose that we have somez such that z ∈ S, and x ∈ Tj . By Pi+1, if h(z) = ±1 then h(z + x) = h(z). Also byPi+1, if h(z) is even, then so is h(z + x). Now, if h(z) �= h(z + x), then z ∈ Tj+1,contradicting our assumption. Thus h(z + x) = h(z) and we can apply Lemma 5.15 toobtain the desired result. �

5.4. Proof that s > 1

Finally, we show that s, the greatest common divisor of the Ti , is greater than 1.

Lemma 5.19. Let G be an arbitrary finite abelian group, which we write additively.Let Y be a set of generators for G, closed under negation. Fix some a ∈ G. For anyb in G, it is possible to write b as the sum of a series of elements from Y, so that noproper partial sum of the series equals a (i.e., excluding the empty partial sum andthe complete partial sum).

Proof. The proof is by induction on |G|. If G is cyclic, pick x ∈ Y a generator for G.If b occurs before a in the sequence x, 2x, . . ., then we are done. Otherwise, use −x.

If G is not cyclic, find a cyclic subgroup H which is a direct summand, and has agenerator x ∈ Y . Let a, b denote the images of a and b in G/H . Apply the induction


hypothesis to G/H . Lifting to G, we obtain a series whose sum differs from b by anelement of H, which we can dispose of as in the cyclic case above. The only problemoccurs if b = a, b �= a, and the series for G/H happens to sum to a. In this case,instead of putting the series obtained for H after the series for G/H , begin with thefirst term from the series for H, followed by the series for G/H , followed by the restof the series for H. �

Example 5.3. Let G�Z/2Z ⊕ Z/3Z. Let Y = {(1, 0), (0, 1), (0, −1)} Let a = (1, 0),b = (1, 1). If we choose H to be the copy of Z/2Z, then the G/H series is (0, 1), theH series is (1, 0), and we can take ((0, 1), (1, 0)) as our desired series.

If we take H to be the copy of Z/3Z, then the G/H series is (1, 0), and the seriesfor H is (0, 1). In this case we cannot just concatenate the two series, because we arein the undesirable situation described above where b = a and the G/H series sums toa. Thus we take the first term of the H series (which in this case happens to be all ofthe H series), followed by the G/H series, followed by the rest of the H series (whichin this case happens to be empty) and we obtain ((0, 1), (1, 0)) as our desired series.

Lemma 5.20. The greatest common divisor s of all the Ti is greater than 1.

Proof. Suppose otherwise. Let i be as small as possible, so that ti divides k. ByLemma 5.8, i > 0. We will now demonstrate that all multiples of ti which are notmultiples of ti−1 must be type 1. However, since elements of Ti are of even type, thiswould force Ti to be empty, and ti = ti−1, a contradiction.

By Ri−1, adding an element of Ti to an element of type 1 not divisible by ti−1yields another element of type 1. Let x be an arbitrary element of Ti which is nota multiple of ti−1. We wish to write x − k as the sum of a series of elements fromT� i−1 such that, if the partial sums are z1, . . . , zm = x − k, then for no l is k + zl

divisible by ti−1. If we can do this, we can conclude that x is type 1.We know that the elements of Ti generate tiZ/ti−1Z, but in fact more is true. By

Lemma 5.16, we know that Ti contains a set of generators and their negatives fortiZ/ti−1Z. We can therefore apply Lemma 5.19, and we are done. �

6. The cone of F -positive symmetric functions

We now consider the set K of all F ∈ � having a nonnegative representation interms of the basis of fundamental quasisymmetric functions, that is,

K ={∑

�

c�F� ∈ � | c� �0 for all �

}. (6.1)

Since K is the intersection of � with the nonnegative orthant of Q (with respect to thebasis {F�}), Kn := K ∩ �n is a polyhedral cone for each n�0. It contains the Schurfunctions s�, � � n, so it has full dimension in �n.


6.1. The generators of Kn

We consider first the minimal generators of the cone Kn, i.e., its one-dimensionalfaces or extreme rays. These include all the Schur functions and, in general, can becharacterized by a condition of being balanced.

We begin by considering the notion of the spread of a quasisymmetric function. For� � �, we denote by

[�, �]� = {� | � � � � �} (6.2)

the lexicographic interval between � and �. For a quasisymmetric function F =∑c�F� ∈ Q, we define the spread of F to be the smallest lexicographic interval

[�, �]� so that c� = 0 whenever � /∈ [�, �]�.For a partition � � n, we let �′ denote the conjugate partition and define the compo-

sition

� := �([n − 1] \ S

(�′)) . (6.3)

Thus if � = 33, then �′ = 222, so [5] \ S(�′) = {1, 3, 5} ⊂ [5] and � = 1221. Notethat � corresponds to the descent set of the tableaux Tr obtained by filling the Ferrersshape � by rows, � corresponds similarly to descent set of the filling Tc by columnsand � � �, with equality if and only if � is n or 1n. In the example above, we have

Tr = 1 2 34 5 6

and Tc = 1 3 52 4 6

.

Also note that � = � if and only if � = �.

Proposition 6.1. The spread of the Schur function s� is the interval [�, �]�.

Proof. Recall that s� = ∑c�F� where c� is the number of standard Young tableaux T

of shape � with � = �(D(T )). Let Tr , respectively, Tc be the standard Young tableauxobtained by filling the Ferrers diagram with shape � by rows, respectively, by columns.As noted, Tr and Tc correspond in this way to � and �. Now for any other tableauxT, let ir be the first index for which ir + 1 is not in the same row as in Tr and letic be the first index for which ic + 1 is not in the same column as in Tc. Then ir isa descent in T but not in Tr and ic is a descent in Tc but not in T, so � ≺ �(D(T ))

≺ �. �

Lemma 6.2. Suppose �, � � n, � �= �, and the spread of s� is a subset of the spread ofs�. Then if s� = ∑

c�F� it follows that c� = 0.

Proof. By assumption, we have � ≺ � � � ≺ �. The first inequality implies �′ ≺ �′, sothere is a minimum index j > 1 so that

�1 + · · · + �j > �1 + · · · + �j . (6.4)


Now if c� �= 0, then there must be a filling T of the shape � with � (D(T )) = �. Thenindices 1, 2, . . . , �1 need to be in the first row of T, indices �1 + 1, . . . , �1 + �2 needto be in the first two rows, etc. However (6.4) indicates this filling will fail at row j.

�

We can now prove the main result of this section.

Theorem 6.3. The Schur functions s� are extreme in the cone K.

Proof. Suppose s� = F1 + F2 with F1, F2 ∈ K. Then Fi = ∑� ai

�s� with

a1� + a2

� = 1 and a1� + a2

� = 0, � �= �. (6.5)

Suppose Fi = ∑ci�F�.

If there is a � �= � with ai� �= 0, then either � � �, � ≺ � or the spread of s� is

a subset of the spread of s�. If there is such a � with � � �, choose one which islexicographically largest. If not, but there is one with � ≺ �, then choose such a �such that � is lexicographically smallest. Otherwise, choose a lexicographically largest� with the spread of s� a subset of the spread of s�. By Proposition 6.1 and Lemma6.2, one of the Fi must have ci

� < 0 or ci� < 0 for the chosen �.

Thus ai� = 0 for � �= � and so both F1 and F2 are multiples of s�, showing s� to

be extreme. �

Note that there are extremes other than the Schur functions. The first one appearswhen n = 4:

s31 + s211 − s22 = F31 + F13 + F211 + F112

is extreme in K4. In K5, there are two such extremes, s311 + s2111 − s221 and s41 +s311 − s32. In K6, there are 23. At present there is no general description of whichcombinations of Schur functions are extreme.

Consequently, we consider next the problem of determining when a quasisymmetricfunction F = ∑

hSFS is an extreme element of the cone K of F-positive symmetricfunctions. (Here we begin indexing by subsets of [n] in place of compositions ofn + 1, where FS = F�(S).) We relate this to a property of the multicollection {S hS },which leads to the notion of fully balanced multicollections of subsets of a finite set.Fully balanced multicollections with nonnegative multiplicities will yield F-positivesymmetric functions, in general, while minimal such collections give rise to extremes.

We say a subset S ⊂ [n] has profile a1, . . . , ak if S consists of maximal consecutivestrings of length a1, . . . , ak in some order. In this case, |S| = a1+· · ·+ak . For example,{2, 3, 5, 7, 8, 9} ⊂ [11] has profile 321000. For � = �1�2 . . . �k � n + 1, define

F� = {S ⊂ [n] | S has profile �1 − 1, . . . , �k − 1}. (6.6)


Thus if S = {2, 3, 5, 7, 8, 9} ⊂ [11], then S ∈ F432111. Further F11...1 = {∅}, S ∈ F21...1if and only if |S| = 1, and S ∈ F221...1 if and only if S = {i, j}, where i < j − 1,while S = {i, i + 1} ∈ F31...1.

We denote a multicollection of subsets of [n] by {S kS | S ⊂ [n]} = {S kS }, wherekS denotes the multiplicity of the subset S. For our purposes, a multicollection {S kS }can have any rational multiplicities kS .

Definition 6.4. Let � � n + 1. A multicollection {S kS } of subsets of [n] is �-balancedif there is a constant �� such that for all T ∈ F�,

∑S⊇T

kS = ��. (6.7)

The multicollection {S kS } is fully balanced if it �-balanced for all � � n + 1.

Multicollections that are 21 . . . 1-balanced have been called balanced in the literatureof cooperative game theory [7], although there the term is applied to the underlyingcollection whenever positive multiplicities kS exist.

Theorem 6.5. A homogeneous quasisymmetric function F = ∑S hSFS ∈ Qn+1 is sym-

metric if and only if the multicollection {S hS } of subsets of [n] is fully balanced.

Proof. Note that, for � � n+ 1, R ∈ F� if and only if � (�([n] \ R)) = �. Further, notethat if T ∈ F� and R ⊂ T , R �= T , then R ∈ F� for some � ≺ �. By inclusion–exclusion, we get

∑S⊇T

hS =∑R⊆T

(−1)|R| ∑S⊆[n]\R

hS =∑R⊆T

(−1)|R|f[n]\R, (6.8)

where fS and hS are related as d� and c� in (2.5). Now, F is symmetric if and onlyif f[n]\R only depends on � for R ∈ F�. Thus if F is symmetric, then (6.8) shows thesum

∑S⊇T hS to depend only on � (and � ≺ �) when T ∈ F�.

Now suppose the multicollection {S hS } is �-balanced for all � � n + 1. We argue byinduction on the lexicographic order on partitions. We assume f[n]\R only depends on� for all R ∈ F�, � ≺ �. (The base case for � = 11 . . . 1 is trivial.) For T ∈ F�, theassertion now follows from (6.8), since the number of R ⊂ T with R ∈ F�, for � ≺ �,depends only on �. �

Thus, elements of Kn+1 correspond to fully balanced collections with nonnegativemultiplicities. Those with minimal support {S | hS �= 0} correspond to the extremesof the cone. One can view integral extremes of Kn+1 as combinatorial designs of anextremely balanced sort: each element of [n] is in the same number of sets (countingmultiplicity), as are each nonadjacent pair, each adjacent pair, etc. One is led to wonder


whether the designs coming this way from Schur functions have special propertiesamong these. The first of these for which the multiplicities are not all one is

s321 = F{1,3} + F{1,4} + F{2,3} + 2F{2,4} + F{2,5} + F{3,4} + F{3,5}+ F{1,2,4} + F{1,2,5} + F{1,3,4} + 2F{1,3,5} + F{1,4,5} + F{2,3,5} + F{2,4,5}.

Here �21111 = 8, �3111 = 2, �2211 = 4, �321 = 1 and �222 = 2.

6.2. The facets of Kn

To describe the facets of Kn, we rewrite (6.1) as follows. Since the Schur functionss�, � � n, are a basis for �n, writing s� = ∑

�[s�]F�F�, we see that

Kn =⎧⎨⎩∑

� � n

c�s� |∑�

c�[s�]F� �0 for all � � n

⎫⎬⎭ . (6.9)

Eq. (6.9) gives 2n−1 inequalities for Kn, one for each � � n. However, when � ∼ �,these inequalities are identical (see Definition 2.3). In fact, we conjecture that theseare the only redundant inequalities, so the facets of Kn would be in bijection with theequivalence classes of compositions under ∼.

The inequality for Kn given by c� �0 in (6.1) is redundant if and only if there exista� �0 such that

c� =∑��∼�

a�c� (6.10)

holds for all F = ∑c�F� ∈ �.

For each composition � � n we define the vector v� = (v�,�; � � n) by v�,�= multM(�)(�). By definition, � ∼ � if and only if v� = v�.

Proposition 6.6. The inequality c� �0 is redundant for some � � n if and only if v� isnot extreme in the convex hull of all v�, � � n.

We end with the following conjecture.

Conjecture 6.1. Any one, and so all, of the equivalent statements holds:

(1) The facets of Kn are in bijection with the equivalence classes of compositions� � n,

(2) The inequalities c� �0, � � n, are all irredundant,(3) Each v� is extreme in the convex hull of all v�, � � n.


One can imagine an approach to Conjecture 6.1 that uses Theorem 4.1 along witha separation argument for v� that targets the decomposition structure of thecomposition �.

6.3. Ribbon-positivity

One can call a symmetric function F-positive if it belongs to the cone K. BeingF-positive is a weakening of the condition of being Schur-positive. Both F-positivityand Schur-positivity are closed under taking products. The quasisymmetric functionsFS have been identified with characters of 0-Hecke algebras [13, §4.1], lending arepresentation-theoretic interpretation to being F-positive.

A strengthening of Schur-positivity would be what we might call ribbon-positivity,that is, belonging to the cone in � spanned by the ribbon Schur functions r�, � � n. By(2.2), this cone is also closed under taking products. As far as we know, this ribboncone has not been studied.

Stronger still would be membership in the cone spanned by the r�, � � n. That thissimplicial cone is strictly smaller than the ribbon cone can be seen from the fact thatr132 = r321 + r33 − r42, which follows from (2.3). This relation suggests that the ribbonSchur functions might all be extreme in the ribbon cone.

We know of one example of ribbon-positive symmetric functions. In [6], Hersh showsthat the chain-enumeration quasisymmetric function of k-shuffle posets are ribbon-positive by showing they are sums of products of ribbon Schur functions.

Acknowledgments

The authors are grateful to Marcelo Aguiar, Nantel Bergeron, Curtis Greene, Alexan-der Postnikov, and Richard Stanley for useful comments over the course of this work,and to GAP [3] for helpful calculations.

References

[1] L.J. Billera, N. Liu, Noncommutative enumeration in graded posets, J. Algebraic Combin. 12 (2000)7–24.

[2] W. Fulton, Young Tableaux, Cambridge University Press, Cambridge, UK, 1997.[3] The GAP Group, GAP—Groups, Algorithms, and Programming, Version 4.3; 2002

(http://www.gap-system.org).[4] I.M. Gel’fand, D. Krob, A. Lascoux, B. Leclerc, V. Retakh, J.-Y. Thibon, Noncommutative symmetric

functions, Adv. Math. 112 (1995) 218–348.[5] I. Gessel, C. Reutenauer, Counting permutations with given cycle structure and descent set,

J. Combin. Theory Ser. A 64 (1993) 189–215.[6] P. Hersh, Chain decomposition and the flag f-vector, J. Combin. Theory Ser. A 103 (2003) 27–52.[7] Y. Kannai, The core and balancedness, in: R.J. Aumann, S. Hart (Eds.), Handbook of Game Theory

with Economic Applications, vol. 1, Elsevier Science Publishers, (North-Holland), Amsterdam, 1992(Chapter 12).

[8] A.N. Kirillov, A. Kuniba, T. Nakanishi, Skew diagram method in spectral decomposition of integrablelattice models, Comm. Math. Phys. 185 (1997) 441–465.

http://www.gap-system.org


[9] A. Lascoux, P. Pragacz, Ribbon Schur functions, European J. Combin. 9 (6) (1988) 561–574.[10] I. Macdonald, Symmetric Functions and Hall Polynomials, second ed., Oxford University Press, New

York, USA, 1995.[11] P.A. MacMahon, Combinatory Analysis, Cambridge University Press, 1917 (vol. I), 1918 (vol. II),

reprint, Chelsea, New York, USA, 1960.[12] R. Stanley, Enumerative Combinatorics, vol. 2, Cambridge Studies in Advanced Mathematics, vol.

62, Cambridge University Press, Cambridge, UK, 1999.[13] J.-Y. Thibon, Lectures on noncommutative symmetric functions, in: Interaction of Combinatorics

and Representation Theory, MSJ Memoirs, vol. 11, Mathematical Society of Japan, Tokyo, 2001,pp. 39–94.

[14] S. van Willigenburg, Equality of Schur and skew Schur functions, Ann. Comb., to appear.

steph/papers/advanceshdl.pdf · advances in mathematics 204 (2006) 204–240 decomposable...

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